# How can I calculate the pixel position in a figure-windows of a 3D object in perspective projection?

13 visualizaciones (últimos 30 días)
J-G van der Toorn el 2 de Mzo. de 2016
Comentada: J-G van der Toorn el 12 de Mzo. de 2019
I have an animation of a complex scene of 3D objects (surfaces and patches) in an axis that has 'projection' set to 'perspective', while my cam(era)pos(ition) and cam(era)target change during the animation. I want to draw a trace of one of the object in the window, as a line connecting the center of this object in the subsequent animation frames. Therefore, I need to calculate the pixel position of that object, and draw that on a (invisible) static plane axis in the same figure window.
I am stuck in calculating the pixel position.
[az,el] = view(axis1);
fov = camva(axis1);
viewmatrix = viewmtx(az,el,fov,campos(axis1)+arbitrary_offset);
% arbitrary_offset is needed to get the correct projection
pixelposition = viewmatrix*[xyz;ones(1,size(xyz,2))];
% xyz is the position in the 3D frame
line(axis2,pixelposition(1)./pixelposition(4),pixelposition(2)./pixelposition(4));
In order to match the calculated pixelposition to the rendered scene, I have to modify the axis scaling of the 2nd axis, the axis limits, and an arbitrary_offset for the position used in the call to viewmtx. And still, it's not perfect, and the offset has to be changed when I resize the figure window.
What I would like is a function that takes the axis parameters of the 3D axis (position, aspect ratio, axis limits, camera parameters like position, target, upvector, viewangle) and the axis limits of the 2D overlay axis, and calculates the correct corresponding 4x4 transformation matrix.
How do I do that? Are there Matlab functions for this purpose available? Or can I request somehow the pixelposition of a point in a 3D axis directly through some hidden property?
##### 1 comentarioMostrar -1 comentarios más antiguosOcultar -1 comentarios más antiguos
J-G van der Toorn el 10 de Mzo. de 2016
I see a main obstacle in getting this solved in a robust way: displaying axes with either PlotBoxAspectRatio or DataAspectRatio set to manual, causes the axes to be resized within the figure, without updating axis' position property. In fact, not a single (observable) property of the scaled axis is changing when the containing figure is being resized. The Position property of the axis is the containing pox, not the actual one. This 3D-to-2D function can only work well when the actual axis position and dimension, as calculated under-water by the Matlab graphics engine, is available.

Iniciar sesión para comentar.

Mike Garrity el 2 de Mzo. de 2016
Editada: Mike Garrity el 2 de Mzo. de 2016
I'm afraid that it's rather more complex than just getting the view matrix.
Let's walk through an example in detail. I'll assume that you're in 3D, with a perspective projection, and 'axis vis3d'.
Create 2 axes with a line each. We'll make axis1 3D, and axis2 2D in pixels. Then we'll try to match the lines.
fpos = get(gcf,'Position');
axis1 = axes;
axis2 = axes('Position',[0 0 1 1], ...
'Visible','off','HandleVisibility','off','PickableParts','none', ...
'XLim',[1 fpos(3)],'YLim',[1 fpos(4)]);
l1 = line(nan,nan,nan,'LineWidth',5,'Color',[.75 .75 .75],'Parent',axis1);
l2 = line(nan,nan,nan,'LineWidth',1,'Color','red','Parent',axis2);
Give l1 random coordinates, and give axis1 a random view.
l1.XData = randn(1,5);
l1.YData = randn(1,5);
l1.ZData = randn(1,5);
view(axis1,360*rand,180*rand-90);
axis(axis1,'vis3d')
axis1.Projection = 'perspective';
Pull coordinates out of line 1 and append W.
xyzw = [l1.XData; l1.YData; l1.ZData; ones(size(l1.XData))];
We need 4 things to convert xyzw to pixels. - Model transform - View transform - Viewport - Projection transform
Compute model transform
xl = axis1.XLim;
yl = axis1.YLim;
zl = axis1.ZLim;
xscale = 1/diff(xl);
yscale = 1/diff(yl);
zscale = 1/diff(zl);
model_xfm = [xscale, 0, 0, -xl(1)*xscale; ...
0, yscale, 0, -yl(1)*yscale; ...
0, 0, zscale, -zl(1)*zscale; ...
0, 0, 0, 1];
Compute view transform
[az,el] = view(axis1);
view_xfm = viewmtx(az,el);
view_offset = ((axis1.CameraPosition-axis1.CameraTarget).*[xscale,yscale,zscale] ...
+ [1/2 1/2 1/2]);
view_offset = view_xfm*[view_offset,1]';
view_xfm(1:3,4) = -view_offset(1:3);
Compute viewport and aspect ratio
old_units = axis1.Units;
axis1.Units = 'pixels';
viewport = axis1.Position;
axis2.Units = old_units;
ar = viewport(3)/viewport(4);
Compute projection transform
fov = axis1.CameraViewAngle;
tanfov = tand(fov/2);
n = .1;
f = 10;
r = tanfov * ar * n;
l = -r;
t = tanfov * n;
b = -t;
proj_xfm = [2*n/(r-l), 0, (r+l)/(r-l), 0; ...
0, 2*n/(t-b), (t+b)/(t-b), 0; ...
0, 0, -(f+n)/(f-n), -2*f*n/(f-n); ...
0, 0, -1, 0];
Now w can compute pixel coordinates.
Multiply xyzw by the 3 transform matrices.
ndc = proj_xfm*view_xfm*model_xfm*xyzw;
% Go from normalized device coordinates [-1 to 1] to pixels.
dc = [.5*(1+ndc(1,:)./ndc(4,:)); .5*(1+ndc(2,:)./ndc(4,:))];
dc(1,:) = viewport(1) + viewport(3)*dc(1,:);
dc(2,:) = viewport(2) + viewport(4)*dc(2,:);
% Place the resulting coordinates on l2.
l2.XData = dc(1,:);
l2.YData = dc(2,:);
l2.ZData = zeros(1,size(dc,2));
As I said, I've made some simplifying assumptions here. In particular, if you're letting the PlotBoxAspectRatio and DataAspectRatio be autocalculated, then you need to account for the factors described on this page of the documentation. But hopefully this will be enough to get you unstuck.
##### 8 comentariosMostrar 6 comentarios más antiguosOcultar 6 comentarios más antiguos
Mike Garrity el 9 de Mzo. de 2016
That's great!
Dmitry el 22 de Mzo. de 2016
Editada: Dmitry el 22 de Mzo. de 2016
Dear J-G van der Toorn, I'm trying to solve very similar problem. I'll be glad if you post your final code on matlabcentral.

Iniciar sesión para comentar.

### Más respuestas (1)

Ulrich Reif el 12 de Mzo. de 2019
ds2fig on FX does a perfect job.
##### 1 comentarioMostrar -1 comentarios más antiguosOcultar -1 comentarios más antiguos
J-G van der Toorn el 12 de Mzo. de 2019

Iniciar sesión para comentar.

### Categorías

Más información sobre Detection and Tracking en Help Center y File Exchange.

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by